A baseball team wins $75\%$ of games, and the team wins $90\%$ of their home games. The team has $50\%$ of the games at home.
a. The team won their next game, so what is the probability that it was a home game?
b. The team lost their next game, what is the probability it was a road game?
So I solved both parts but I am not sure I am correct. Since there is a $50\%$ chance the teams games are home then there is a $50\%$ chance the game is away, which is pretty easy.
Then if the game is home they have a $90\%$ chance to win and a $10\%$ chance to lose. If the game is away they have a $75\%$ chance to win and a $25\%$ chance to lose.
For part a) I did $P(Home~and~Win)=(.5)(.9)$ and for part b) $P(Away ~and ~Loss)=(.5)(.25)$. I am on the right track or am I looking at this problem all wrong…
Best Answer
First, if the game is away they have a greater than $25\%$ chance to lose. They lose very few games at home, so must lose a bunch on the road. You need to use the fact that half the games are home.
Then you asked for conditional probabilities. Given that they win, you have the chances of (home and win) and (away and win). The chance the win was at home is $\frac {home\ and\ win}{win}$